Second partial derivative of v=e^(x*e^y)

[sdoyle1]

Homework Statement

Find the second partial derivative of v=e^(x*e^y)

Homework Equations

I know that I need to find Vx and Vy first and then the second partial derivative would be Vxx, Vyy, Vxy.

The Attempt at a Solution

I'm really confused on how to find Vx or Vy
Vx= the derivative with regards to x, if y is a constant
so would it be Vx=e^(x*e^y) * (e^y)?
Any help would be great

 

[gb7nash]

Find the second partial derivative of v=e^(x*e^y)

I know that I need to find Vx and Vy first and then the second partial derivative would be Vxx, Vyy, Vxy.

You're forgetting one.

Vx=e^(x*e^y) * (e^y)

That's correct. What's Vy?

 

[sdoyle1]

Would Vy= e^(x*e^y) * (xe^y)

 

[gb7nash]

Would Vy= e^(x*e^y) * (xe^y)

Correct. Now you need to take the second partial derivatives.

 

[sdoyle1]

Vxx would be the second derivative with respect to x but keeping y as a constant. This is where I get confused. Would it be:

Vxx=e^(xye^y)(ye^y)
= e^(xy^2e^y)

 

[sdoyle1]

Wait, you would add the exponents, not multiply them.
So Vx= e^((xe^y)+y)
Vxx= e^(xe^y+y) * (e^y)

 

[gb7nash]

Vxx= e^(xe^y+y) * (e^y)

Correct

 

[sdoyle1]

Ok then Vyy= xe^((xe^y)+y) * (x(e^y) +1)

Then Vxy= e^((xe^y)+y) * ((xe^y)+1) and Vyx is the same as Vxy

As an aside, how would I integrate t(t-1)^1/2 ?

 

[gb7nash]

Ok then Vyy= xe^((xe^y)+y) * (x(e^y) +1)

Then Vxy= e^((xe^y)+y) * ((xe^y)+1) and Vyx is the same as Vxy

Correct.

As an aside, how would I integrate t(t-1)^1/2 ?

Try integration by parts.

 

[SammyS]

...

As an aside, how would I integrate t(t-1)^1/2 ?

Use the substitution: u = t-1 .

 

[gb7nash]

Good call. Both methods should work, but this is much simpler to do.